Fit an exponential function to time-series data

Hi All,

I have a time series data and I would like to fit an exponential curve using the following expression to the data points and estimate the relaxation time B

The dataset can be found here.

I saw some examples of curve fitting here https://in.mathworks.com/help/matlab/math/example-curve-fitting-via-optimization.html . However it is not clear to me how to fit the above function and estimate B. Could someone please help me with this? I would like to find time scale at which C reaches 100 % of C0, C0 = 7.5.

Side note, I've the optimization toolbox and curve fitting tool box installed.

Answer 1

[Mathieu NOE]

hello

this is the poor man solution without any toolbox

data was stored in txt file

your B = -1/b_sol from my code below

a_sol = 8.3470

b_sol = -0.7241

c_sol = 0.3345

d_sol = -0.8432

numericData = importdata('data.txt');
start = 25; % use data only starting at this index
x  = numericData(start:end,1);
y2fit = numericData(start:end,2);
% exponential fit method 
% code is giving good results with template equation : % y = a.*(1-exp(b.*(x-c))) + d;

f = @(a,b,c,d,x) a.*(1-exp(b.*(x-c))) + d;
obj_fun = @(params) norm(f(params(1), params(2), params(3), params(4),x)-y2fit);
sol = fminsearch(obj_fun, [y2fit(end)-y2fit(1),0,0,y2fit(1)]);
a_sol = sol(1)
b_sol = sol(2)
c_sol = sol(3)
d_sol = sol(4)
y_fit = f(a_sol, b_sol,c_sol ,d_sol, x);

figure
plot(x,y2fit,'r',x,y_fit,'-.k');
legend('data','exp fit');

[Deepa Maheshvare]

@Mathieu NOE Thanks so much. Could you please explain a bit on how B = -1/b_sol ?

[Deepa Maheshvare]

@Mathieu NOE 1/0.72 gives 1.38s. I am not sure how to releate this to the time scale at which steady state is achieved. Because from the plot, I find that steady-state is reached after 5seconds. Could you please offer some explanation on this too?

[Mathieu NOE]

hello again

1/ you see that my fit function has not exactly the same form as yours

I used x as the time variable and the "b" are used differently :

my function : y = a.*(1-exp(b.*(x-c))) + d ;

your function : = a.*(1-exp(-x/B)) ;

we don't care about c and d here , and it's quite obvious now that : B (your formula) = -1/b (mine, and solution in the code once the fit is done , is b_sol);

2/ B is a decay rate - see definition here : Exponential decay - Wikipedia

a very often used relationship : to reach 95% of the asymptote, it takes t = 3 x decay rate = 3 * 1.38 = 4.14 s

theoretically, you never reach 100% (the asymptote) , you can only get very close